University Physics II · topic UPII-04.02
Electric potential and potential difference
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Electric potential and potential difference: predict a finite change
The electric potential at a point is the electric potential energy per unit charge that a small test charge would have there, measured in volts, where one volt is one joule per coulomb. Use the supplied topic model Y_out = -ef*xp/100, where Y_out is V(P) − V(LEFT PLATE) (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. xp is Probe position x_P, measured in cm. Keep ef (Field E from left plate to right) = 1200 V m⁻¹; xz (Position where V is set to zero) = 5 cm fixed. For this exercise use only 1.50937 cm ≤ xp ≤ 8.99062 cm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at xp = 3.15525 cm and 4.20262 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δxp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
V(P) − V(LEFT PLATE) against xp
Predictions from the supplied Electric potential and potential difference model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): V(P) − V(LEFT PLATE) / V
Horizontal axis (x): xp: Probe position x_P / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xp: Probe position x_P / cm | V(P) − V(LEFT PLATE) / V |
|---|---|
| 1.509375 | -18.1125 |
| 3.3796874999999997 | -40.55625 |
| 5.25 | -63 |
| 7.120312499999999 | -85.44374999999998 |
| 8.990625 | -107.8875 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of xp separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At xp = 3.15525 cm, substitute into Y_out = -ef*xp/100: Y₀ = -37.863 V.
- At xp = 4.20262 cm, the same fixed-parameter model gives Y₁ = -50.4315 V.
- Signed change: ΔY_out = Y₁ − Y₀ = -12.5685 V; the input change is Δxp = 1.04737 cm.
- Average sensitivity is ΔY_out/Δxp = -12 V per cm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = -37.863 V, Y₁ = -50.4315 V; ΔY_out = -12.5685 V; average sensitivity = -12 V per cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Electric potential and potential difference: infer an input with uncertainty
The electric potential at a point is the electric potential energy per unit charge that a small test charge would have there, measured in volts, where one volt is one joule per coulomb. Use the supplied topic model Y_out = -ef*xp/100, where Y_out is V(P) − V(LEFT PLATE) (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. xp is Probe position x_P, measured in cm. Keep ef (Field E from left plate to right) = 1200 V m⁻¹; xz (Position where V is set to zero) = 5 cm fixed. For this exercise use only 1.50937 cm ≤ xp ≤ 8.99062 cm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -72.87525 V, with a bounded reading uncertainty ±5.2518 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown xp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 cm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
V(P) − V(LEFT PLATE) against xp
Predictions from the supplied Electric potential and potential difference model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): V(P) − V(LEFT PLATE) / V
Horizontal axis (x): xp: Probe position x_P / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xp: Probe position x_P / cm | V(P) − V(LEFT PLATE) / V |
|---|---|
| 1.509375 | -18.1125 |
| 3.3796874999999997 | -40.55625 |
| 5.25 | -63 |
| 7.120312499999999 | -85.44374999999998 |
| 8.990625 | -107.8875 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(xp) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.50937) = -18.1125 V and Y_out(8.99062) = -107.888 V. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives xp_line = 1.50937 + (-72.8752 − -18.1125)(8.99062 − 1.50937)/(-107.888 − -18.1125) = 6.07294 cm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives xp = 6.07294 cm; substituting back gives Y_out = -72.87525 V.
- The interpolation error in the input is |xp_line − xp| = 1.77636e-15 cm. This is within the requested numerical tolerance.
- Invert both limits Y_out = -78.12705 V and -67.62345 V. Sort the roots even if the response decreases: 5.63529 cm ≤ xp ≤ 6.51059 cm.
- The lower and upper input deviations are 0.43765 cm and 0.43765 cm. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer xp ≈ 6.07294 cm; compatible interval [5.63529, 6.51059] cm. Endpoint interpolation gives 6.07294 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.